Question

Cyclohexane and ethanol at a particular temperature have vapour pressures of $280mmHg$ and $168mmHg$ respectively. If the mole fraction of cyclohexane in the solution mixture is 0.32, the mixture is found to have a total vapour pressure of $250mmHg$.

Choose the correct statement(s).

• A. The solution mixture obeys Raoult's law
• B. The solution mixture shows positive deviation from Raoult's law
• C. The solution mixture shows negative deviation from Raoult's law
• D. The solution mixture is an non-Ideal solution

Answer 1

Answer: (B), (D)

The solution mixture is an non-Ideal solution

Given,
$\text{Pure vapour pressure of cyclohexane,}{p}_A^{\circ }=280mmHg\text{Pure vapour pressure of ethanol,}{p}_A^{\circ }=168mmHg$

$\text{Mole fraction of cyclohexane,}{X}_A=0.32\therefore \text{Mole fraction of ethanol,}{X}_B=1-0.32=0.68$

By Raoult's law
Total vapour pressure of solution,
$P_T=X_A.p_A^{\circ }+X_B.p_B^{\circ }$
$P_T=280\times 0.32+168\times 0.68$
$=203.84mmHg$
For the solution to be an ideal one, the vapour pressure should be $203.84mm$ as calculated from Raoult's law but the given value of vapour pressure is $250mm$, so the solution is not ideal. It violates the Raoult's law. Hence, it is an non-Ideal solution.
In given solution mixture,
$P_T>X_A.p_A^{\circ }+X_B.p_B^{\circ }$
$250mmHg>203.84mmHg$
Thus, the given mixture shows positive deviation from the Raoult's law.
Cyclohexane-Ethanol interaction is less than the Ethanol-Ethanol and the Cyclohexane-Cyclohexane interaction.